Question

The standard cell potential, \(E^{\circ},\) for the reaction of \(\mathrm{Zn}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{~g})\) is \(2.12 \mathrm{~V}\). Write the chemical equation for the reaction of \(1 \mathrm{~mol}\) zinc. Calculate the standard Gibbs free energy change, \(\Delta_{t} G^{\circ},\) for this reaction.

Short answer

The standard Gibbs free energy change, \( \Delta_{t} G^{\circ}, \) is \( -408.75 \mathrm{~kJ/mol} \).

Step-by-step solution

Identify the Half-Reactions

To find the overall reaction involving Zn and Cl₂, let's first write the half-reactions. Zinc undergoes oxidation: \[ \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \] and chlorine undergoes reduction: \[ \mathrm{Cl}_{2}(g) + 2e^- \rightarrow 2\mathrm{Cl}^-(aq) \].

Write the Balanced Equation

Combine the oxidation and reduction half-reactions to balance the electrons and obtain the overall cell reaction: \[ \mathrm{Zn}(s) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Zn}^{2+}(aq) + 2\mathrm{Cl}^-(aq) \]. This represents the reaction of 1 mole of zinc metal with chlorine gas.

Relate Standard Cell Potential to Gibbs Free Energy

Use the formula \( \Delta_{t} G^{\circ} = -nFE^{\circ} \) where \( n \) is the number of moles of electrons transferred (which is 2 in this reaction), \( F \) is Faraday's constant (approximately \( 96485 \) C/mol), and \( E^{\circ} \) is the standard cell potential.

Substitute Values into the Formula

Substitute the given values into \( \Delta_{t} G^{\circ} = -nFE^{\circ} \): \[ \Delta_{t} G^{\circ} = -(2)(96485 \mathrm{~C/mol})(2.12 \mathrm{~V}) \].

Calculate Gibbs Free Energy Change

Perform the multiplication to find \( \Delta_{t} G^{\circ} \): \( \Delta_{t} G^{\circ} = -408752.8 \mathrm{~J/mol} \), which is often expressed in kilojoules, so \( \Delta_{t} G^{\circ} = -408.75 \mathrm{~kJ/mol} \).

Key concepts

Gibbs Free Energy

Gibbs Free Energy, commonly denoted as \( \Delta G \), is a thermodynamic quantity that represents the maximum reversible work that a thermodynamic system can perform. It is essential in predicting whether a process will occur spontaneously.
In the context of electrochemistry, Gibbs Free Energy change, \( \Delta_t G^{\circ} \), is related to the cell potential and the amount of charge transferred in the reaction.

• A negative \( \Delta G \) indicates a spontaneous reaction, meaning it can occur without outside intervention.
• A positive \( \Delta G \) means the reaction requires energy input to proceed.

The formula \( \Delta_t G^{\circ} = -nFE^{\circ} \) links Gibbs Free Energy with the standard cell potential \( E^{\circ} \), the number of moles of electrons \( n \), and Faraday's constant \( F \). Here, \( F \) is approximately \( 96485 \) Coulombs per mole. This relation shows that the energy change depends on both the cell potential and how many electrons are moving in the reaction.

Standard Cell Potential

The Standard Cell Potential, \( E^{\circ} \), provides insights into the driving force of an electrochemical reaction. It's essentially about the voltage or electromotive force of a cell when concentrations of all reactants and products are at standard conditions, typically \( 1 \) M concentrations, \( 1 \) atm pressure, and \( 25 \degree C \).

• It is measured in volts (V).
• An electrochemical cell with a positive \( E^{\circ} \) tends to drive the reaction spontaneously.
• Conversely, a negative \( E^{\circ} \) implies a non-spontaneous reaction needing external power to occur.

When calculating the standard cell potential for a reaction involving zinc and chlorine gas, as given \( 2.12 \) V, it tells us that the reaction is spontaneous and can proceed without energy input, at standard conditions. The larger the positive value, the more tendency the reaction has to occur.

Redox Reactions

Redox reactions are chemical processes in which there is a transfer of electrons between two species. The term 'redox' comes from "reduction-oxidation" reactions. It involves two half-reactions:

• Oxidation: Loss of electrons. In the example, \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \), zinc is losing electrons.
• Reduction: Gain of electrons. The chlorine gas \( \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \), is gaining those electrons.

These half-reactions occur simultaneously, where one species is oxidized and another is reduced.
Understanding these reactions is vital in fields like electrochemistry, as they drive the generation of electricity in galvanic cells. In the given reaction of zinc with chlorine, zinc acts as the reducing agent while chlorine is the oxidizing agent.